The value of `C_(P) - C_(v) = 1.00 R` for a gas in state `A` and `C_(P) - C_(v) = 1.06 R` in another state. If `P_(A)` and `P_(B)` denote the pressure and `T_(A)` and `T_(B)` denote the temperature in the two states, then

Options

A. `P_(A) = P_(B), T_(A) gt T_(B)`
B. `P_(A) lt P_(B), T_(A) = T_(B)`
C. `P_(A) lt P_(B), T_(A) gt T_(B)`
D. `P_(A) = P_(B), T_(A) lt T_(B)`

Explanation

To solve the problem, we need to analyze the relationship between the specific heats \( C_P \) and \( C_V \) in two different states of a gas, as well as how these relate to pressure and temperature.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- For state A: \( C_P - C_V = 1.00 R \)
- For state B: \( C_P - C_V = 1.06 R \)

2. **Using the Relationship of Specific Heats:**
- The difference between the specific heats can be expressed as:
\[ C_P - C_V = nR \]
- Therefore, for state A:
\[ n_A R = 1.00 R \implies n_A = 1.00 \]
- And for state B:
\[ n_B R = 1.06 R \implies n_B = 1.06 \]

3. **Applying the Ideal Gas Law:**
- The ideal gas law states:
\[ PV = nRT \]
- From this, we can express pressure \( P \) in terms of \( n \) and \( T \):
\[ P = \frac{nRT}{V} \]
- Since \( V \) (volume) is constant for both states, we can say:
\[ P_A \propto n_A T_A \quad \text{and} \quad P_B \propto n_B T_B \]

4. **Comparing Pressures:**
- Since \( n_A < n_B \) (1.00 < 1.06), we can infer that:
\[ P_A < P_B \]

5. **Comparing Temperatures:**
- From the relationship \( T \) is inversely proportional to \( n \) (assuming constant pressure):
\[ T_A \propto \frac{1}{n_A} \quad \text{and} \quad T_B \propto \frac{1}{n_B} \]
- Since \( n_A < n_B \), it follows that:
\[ T_A > T_B \]

### Conclusion:
- We conclude that:
- \( P_A < P_B \)
- \( T_A > T_B \)

### Final Answer:
The correct relationships are:
- Pressure: \( P_A < P_B \)
- Temperature: \( T_A > T_B \)

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